The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use, \( g : 10 \, \text{m/s}^2 \)]
- \( 6\sqrt{5} \, \text{ms}^{-1} \)
- \( 5\sqrt{6} \, \text{ms}^{-1} \)
- \( 5\sqrt{5} \, \text{ms}^{-1} \)
- \( 2\sqrt{5} \, \text{ms}^{-1} \)
The Correct Option is A
Approach Solution - 1
To solve this problem, we need to determine the speed of the pendulum bob at the lowest point after considering the energy dissipated due to air resistance.
Initially, when the bob is released from the horizontal position, its entire energy is potential energy given by:
\[ E_{\text{initial}} = m \cdot g \cdot h \]
where:
- \(m\) is the mass of the bob (we will see that it cancels out later)
- \(g = 10 \, \text{m/s}^2\) is the acceleration due to gravity
- \(h = 10 \, \text{m}\) is the height
Therefore, the initial energy is:
\[ E_{\text{initial}} = m \cdot 10 \cdot 10 = 100m \, \text{J} \]
According to the problem, 10% of the initial energy is dissipated, which implies 90% of the initial energy is converted to kinetic energy at the lowest point. Thus, the energy at the lowest point is:
\[ E_{\text{kinetic}} = 0.9 \times 100m = 90m \, \text{J} \]
The kinetic energy of the bob at the lowest point is given by:
\[ E_{\text{kinetic}} = \frac{1}{2} m v^2 \]
Setting the kinetic energy equal to the available energy, we have:
\[ \frac{1}{2} m v^2 = 90m \]
Cancelling \(m\) from both sides:
\[ \frac{1}{2} v^2 = 90 \]
Solving for \(v\) by multiplying both sides by 2:
\[ v^2 = 180 \]
Taking the square root of both sides, we find:
\[ v = \sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5} \, \text{ms}^{-1} \]
Therefore, the speed with which the bob arrives at the lowest point is \(6\sqrt{5} \, \text{ms}^{-1}\). Hence, the correct answer is:
\( 6\sqrt{5} \, \text{ms}^{-1} \)
Approach Solution -2
Given:
- Length of the pendulum, \( \ell = 10 \, \text{m} \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)
- 10% of the initial energy is dissipated against air resistance.
Step 1. Calculate the initial energy:
Since the bob is released from a horizontal position, the initial potential energy (at the top) is:
\(\text{Initial energy} = mg\ell\)
Step 2. Calculate the energy at the lowest point:
Since 10% of the initial energy is dissipated, only 90% of the initial energy is available at the lowest point.
\(\text{Energy at the lowest point} = \frac{9}{10} mg\ell\)
Step 3. Relate energy to speed at the lowest point:
At the lowest point, this energy is entirely kinetic, so:
\(\frac{9}{10}mg\ell = \frac{1}{2}mv^2\)
Simplify by canceling \( m \) from both sides:
\(\frac{9}{10} \times 10 \times 10 = \frac{1}{2}v^2\)
\(v^2 = 180\)
\(v = \sqrt{180} = 6\sqrt{5} \, \text{m/s}\)
Thus, the speed of the bob at the lowest point is \( 6\sqrt{5} \, \text{m/s} \).
The Correct Answer is: \( 6\sqrt{5} \, \text{m/s} \)
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